I want to know how to convert these exponential forms to tensor products of well known logic gates (like the ones built into Qiskit). My goal is to program the Beam-splitter-Kerr ansatz circuit for use in a VQE.
4
-
$\begingroup$ These gates don’t act in a qubit (two-level) Hilbert space, but a qumode (infinite-dimensional) Hilbert space. $\endgroup$– Jahan ClaesApr 24 at 1:03
-
$\begingroup$ @JahanClaes Thanks for pointing that out. In this case could these operators could be projected into the qubit Hilbert space? In this article: arxiv-vanity.com/papers/2103.15021 they use these operators as ansatz for the Bose-Hubbard model. I want to do the same. Is this possible? $\endgroup$– user22395Apr 24 at 4:39
-
$\begingroup$ I don't think that's what you want to do if you're trying to simulate the Bose-Hubbard model. The Bose-Hubbard model is defined on a collection of qumodes, so you don't want to convert the model into qubits. $\endgroup$– Jahan ClaesApr 24 at 19:03
-
$\begingroup$ @JahanClaes Actually I am technically not simulating the Bose-Hubbard model, because I only want to consider the |0> and |1> states. I am not sure the proper name for this - I only referred to it as the Bose-Hubbard model because it is not antisymmetric like the Fermi-Hubbard model. Do you know if there is a name for this Hamiltonian? $\endgroup$– user22395Apr 24 at 19:22
Add a comment
|
3 Answers
$\begingroup$
$\endgroup$
Beam-splitter and Kerr gates are gates which act on infinite-dimensional Hilbert space (qumodes). It is best to use some photonic/bosonic simulator like Strawberry fields (SF) or Bosonic QISKIT to do this. For example, in SF, you will have something like:
BSgate($\theta$, $\pi$) | (q[0], q[1])
Kgate($\cdot$) | q[0]
Kgate($\cdot$) | q[1]
where $\cdot$ is the argument of Kerr gate.
$\begingroup$
$\endgroup$
If one uses the dual-rail encoding for photons (a photon in the "left" waveguide is 0, a photon in the "right" waveguide is 1), then the "gate" of the beam-splitter if just its unitary matrix- ((t, -r)(r, t)). For a 50-50 beam-splitter this is almost equivalent to the Hadamard gate: BS(t=r) = XH.
I am not familiar with the Kerr gate, but I guess you can do the same analogy.
$\begingroup$
$\endgroup$
On a side note, you can give Perceval by Quandela a try, a framework for photonic quantum computing it has native beam splitters, phase shifters, etc and the documentation covers using dual-rail encoding (a.k.a path encoding) but no kerr since it not a linear object.
There are different conventions for beam splitters, but as yaron mentioned, the most common one is the 50-50 BS, which is a Hadamard gate.
As for the kerr, using "Nonlinear Kerr media" section for Nielsen and Chunang, They build a Fredkin such that Fredkin= B†KB, (B: beamsplitter, K: kerr), so do the math, and you will find out how to obtain a simulation of a kerr using Fredkin and Hadamard gates (replacement of the B).
(Fredkin gate is just made of toffoli and Cnot gates)
It does not make much sense to simulate the low-level components that are used to build logical gates by expressing them as logical gates...also creating Nonlinear kerr media is still a challenge in physics... even for major players in photonic quantum computing...
